Which statements describe acceleration? Check all that apply.

As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases

frozen [14]

Bohr's equation for the change in energy is
$\Delta E= \frac{hc}{\lambda}$
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
TRUE

b. As λ increases, f increases and ΔE increases.
FALSE

c. As λ increases, f increases and ΔE decreases.
FALSE

Answer:
As the wavelength increases, the frequency decreases and energy decreases.

3 0

4 years ago

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13. An 80.0 kg climber with a 20.0 kg pack climbs 8848 m to the top of Mount Everest. What is the climber's

miss Akunina [59]

The answer to your question is B

7 0

3 years ago

The headlights of a typical truck are powered by a 12 V battery. What current do they draw if the resistance is 4.8 ohms?

vodomira [7]

Solve this using Olm's law, which relates current (C), voltage (V), and resistance (R). Olm's law says:
$I \: (current ) ={\frac {V \: (voltage) }{R\:(resistance)}}$

We are told that voltage, V = 12V, and resistance, R = 4.8 <span>Ω. Plug these values into the equation and solve for current:
</span> $I = \frac{V}{R}\\
I = \frac{12}{4.8}\\
I = 2.5A

$
<span>-----
Answer: Current = 2.5 Amperes</span>

6 0

4 years ago

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Which of the following describes the magnetic field produced by a current carrying wire? Assume the wire is normal to the page,

Maslowich

D) the magnetic field surrounds the wire like a tube , with a counterclockwise field direction...
as it is in the left hand thumb rule

6 0

4 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$